What is the physical interpretation of the Fourier transform $(\mathcal{F}Z)(t)$ an impedance?

Question

If I compose a impedances out of smaller ones in series and parallel configurations, e.g.

$$Z(\omega)=i\omega L_2+\tfrac{1}{\tfrac{1}{R_1}\ +\ i\omega C_1+\ \tfrac{1}{i\omega L_2}},$$

then I get a an object with functional dependence of the external frequency $\omega$.

One usually consistes signals, how they interact with the frequency depended system they encounter and what one is interested in the output quantities one gets as a result.

Now I wonder:

What is the physical interpretation of the Fourier (or Laplace) transform $(\mathcal{F}Z)(t)$ of this object?

That it has applications is explained in this wikipedia article on the Laplace transform here. There they tell us how, appearently, a dual circuit looks like. It's a little abstract. If I take $Z$ as a function of frequency in above, I don't know what to make of this time dependend object.

Thoguths: First I assume that $Z(\omega)$ will enter the transfer function in some direct way, but I looked it up again and it's not so simple. I see that quantity is merely defined in terms of the input- and output signal, it's not directly a function of the setup, the resistor chains etc., which affect that signal.

In some sense it will certainly tell me how subparts of the circuit react to the frequency of the external voltage. For exmaple by the law of conservation of current "$I_\text{tot}=\sum_i I_i$" we know that this $Z(\omega)$ must affect the voltage drops on single resistors. So since Ohms law relates voltage via resistence to current in a multiplicative way and this current is further diveded into the parts of the circuit by the above relation, I figured the Fourertransform will pop up in a response quantity via the convolutions, because $\mathcal{F}\{f * g\} \propto \mathcal{F}\{f\}\cdot \mathcal{F}\{g\}$.

Answer 1

For concreteness, assume a linear circuit element or network with driving point impedance $Z_a$:

$V_a(j \omega) = Z_a(j \omega) \cdot I_a(j \omega)$

Let:

$I_a(j \omega) = 1 \Leftrightarrow i_a(t) = \delta(t)$.

Now, it is easy to see that the $z_a(t)$ is the time domain voltage response (divided by 1A) due to a current impulse.

This is analogous to the impulse response of a system, $h(t)$, in linear time invariant (LTI) systems theory.

Answer 2

In dynamics of electron beams in particle accelerators or synchrotrons, we describe the self interaction of the beam via an impedance. The impedance interacts with the Fourier transform of the bunch distribution.

In this context, the Fourier transform of the impedance is called the Wake function. It is the explicit force function which gives how one element of the beam interactions with a test particle trailing behind. See, e.g. this textbook by A. Chao http://www.slac.stanford.edu/~achao/wileybook.html for more details.

I don't know if there is a similar interpretation for a circuit, but I suppose it is an interaction between charge moving in one part of the circuit and that in another.